It is easy to verify from the definition for each non-zero square $a$ in $GF(q)$ and each $b$ in $GF(q)$, the each map $$ t_{a,b}: x \mapsto ax+b $$ is an automorphism of the Paley graph. Suppose $q=p^d$ where $p$ is prime. Then the Frobenius map $x\mapsto x^p$ is an automorphism of $GF(q)$ with order $d$, and this is also an automorphism of the Paley graph. Combining all this we get a group of automorphisms of order $dq(q-1)$.
Proving that this is the entire automorphism group is difficult. One of the first proofs is in Carlitz, L. "A theorem on permutations in a finite field". Proc. Amer. Math. Soc. 11 (1960) 456–459. Even the case when $q$ itself is prime is non-trivial. The obvious approach is to use the theorem that a transitive group of prime degree which is not 2-transitive is solvable, and then apply a theorem due to Galois that a solvable group of prime degree consists of translations, as above.